Integraal van $$$-5 + \frac{1}{x^{2}}$$$

Bepaal $$$\int \left(-5 + \frac{1}{x^{2}}\right)\, dx$$$.

Integreer termgewijs:

$${\color{red}{\int{\left(-5 + \frac{1}{x^{2}}\right)d x}}} = {\color{red}{\left(- \int{5 d x} + \int{\frac{1}{x^{2}} d x}\right)}}$$

Pas de constantenregel $$$\int c\, dx = c x$$$ toe met $$$c=5$$$:

$$\int{\frac{1}{x^{2}} d x} - {\color{red}{\int{5 d x}}} = \int{\frac{1}{x^{2}} d x} - {\color{red}{\left(5 x\right)}}$$

Pas de machtsregel $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ toe met $$$n=-2$$$:

$$- 5 x + {\color{red}{\int{\frac{1}{x^{2}} d x}}}=- 5 x + {\color{red}{\int{x^{-2} d x}}}=- 5 x + {\color{red}{\frac{x^{-2 + 1}}{-2 + 1}}}=- 5 x + {\color{red}{\left(- x^{-1}\right)}}=- 5 x + {\color{red}{\left(- \frac{1}{x}\right)}}$$

Dus,

$$\int{\left(-5 + \frac{1}{x^{2}}\right)d x} = - 5 x - \frac{1}{x}$$

Voeg de integratieconstante toe:

$$\int{\left(-5 + \frac{1}{x^{2}}\right)d x} = - 5 x - \frac{1}{x}+C$$

$$$\int \left(-5 + \frac{1}{x^{2}}\right)\, dx = \left(- 5 x - \frac{1}{x}\right) + C$$$A